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P4009 汽车加油行驶问题

P4009 汽车加油行驶问题

题目

题目链接

解法

分层图跑dijkstra。

AC代码

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/**
* @file: P4009.cpp
* @author: yaoxi-std
* @url: https://www.luogu.com.cn/problem/P4009
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
using pii = pair<int, int>;
const int MAXN = 150;
const int MAXK = 15;
const int MAXV = 2e5 + 10;
const int MAXE = 4e6 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
const int DX[4] = {-1, 0, 0, 1};
const int DY[4] = {0, -1, 1, 0};
struct Edge {
int v, cost;
} edge[MAXE];
int n, k, a, b, c, mp[MAXN][MAXN];
int s, t, num, pt[MAXN][MAXN][MAXK];
int tot, head[MAXV], nxt[MAXE], dis[MAXV];
void addedge(int u, int v, int cost) {
edge[++tot] = {v, cost};
nxt[tot] = head[u], head[u] = tot;
}
int dijkstra(int s, int t) {
priority_queue<pii, vector<pii>, greater<pii>> que;
fill(dis, dis + MAXV, INF);
dis[s] = 0, que.push({dis[s], s});
while (!que.empty()) {
int u = que.top().second, d = que.top().first;
que.pop();
if (d > dis[u])
continue;
for (int i = head[u]; i; i = nxt[i]) {
int v = edge[i].v;
if (dis[v] > dis[u] + edge[i].cost) {
dis[v] = dis[u] + edge[i].cost;
que.push({dis[v], v});
}
}
}
return dis[t];
}
signed main() {
read(n), read(k), read(a), read(b), read(c);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
read(mp[i][j]);
for (int l = 0; l <= k; ++l)
pt[i][j][l] = ++num;
}
}
s = pt[1][1][k], t = ++num;
for (int i = 0; i <= k; ++i)
addedge(pt[n][n][i], t, 0);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (mp[i][j]) {
for (int l = 0; l <= k; ++l)
addedge(pt[i][j][l], pt[i][j][k], a);
for (int d = 0; d < 4; ++d) {
int x = i + DX[d], y = j + DY[d];
if (x < 1 || y < 1 || x > n || y > n)
continue;
addedge(pt[i][j][k], pt[x][y][k - 1], d > 1 ? 0 : b);
}
} else {
for (int d = 0; d < 4; ++d) {
int x = i + DX[d], y = j + DY[d];
if (x < 1 || y < 1 || x > n || y > n)
continue;
for (int l = 1; l <= k; ++l)
addedge(pt[i][j][l], pt[x][y][l - 1], d > 1 ? 0 : b);
}
}
addedge(pt[i][j][0], pt[i][j][k], a + c);
}
}
write(dijkstra(s, t)), putchar('\n');
return 0;
}